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 A354411 a(n) is the least oblong number that is divisible by the first n primes. 0
 2, 6, 30, 210, 43890, 510510, 510510, 3967173210, 134748093480, 530514844860, 4201942828713630, 1706257740074998110, 125050509312845636520, 511284700554162118403820, 2695009287439086535873235280, 206794067314254446263154097180, 86753583273488685998907289046220 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 LINKS FORMULA From Michael S. Branicky, May 25 2022: (Start) a(n) <= (m-1)*m, where m = A002110(n). a(n) = m*(m+1), where m = A344005(A002110(n)). (End) a(n) = A118478(n)*(A118478(n)+1). - Chai Wah Wu, May 31 2022 EXAMPLE 2, 3, and 5 are the first three primes. The first oblong number that is a multiple of all three first primes is 30. So, a(3) = 30. The first oblong number that is a multiple of primorial(5) = 2310 is 19*2310 = 43890, so a(5) = 43890. PROG (Python) from sympy import integer_nthroot, primorial def oblong(n): r = integer_nthroot(n, 2)[0]; return r*(r+1) == n def a(n): m = psharp = primorial(n) while not oblong(m): m += psharp return m print([a(n) for n in range(1, 11)]) # Michael S. Branicky, May 25 2022 (Python) # faster alternative using Python 3.8+ A344005(n) by Chai Wah Wu from sympy import primorial def a(n): return (m := A344005(primorial(n)))*(m+1) print([a(n) for n in range(1, 18)]) # Michael S. Branicky, May 26 2022 (PARI) a002110(n) = prod(i=1, n, prime(i)) \\ after Washington Bomfim in A002110 a344005(n) = for(m=1, oo, if((m*(m+1))%n==0, return(m))) a(n) = my(m=a344005(a002110(n))); m*(m+1) \\ Felix Fröhlich, May 31 2022 CROSSREFS Cf. A000040, A002378, A002110, A118478, A344005. Sequence in context: A101178 A294925 A091456 * A293756 A161620 A333508 Adjacent sequences: A354408 A354409 A354410 * A354412 A354413 A354414 KEYWORD nonn AUTHOR Ali Sada, May 25 2022 EXTENSIONS a(9)-a(17) from Michael S. Branicky, May 26 2022 STATUS approved

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Last modified February 6 23:12 EST 2023. Contains 360111 sequences. (Running on oeis4.)